## MATH 221 Lecture 16

Last update: 6 August 2012

## Lecture 16

Example:$\phantom{\rule{1em}{0ex}}$ $\text{Graph}\phantom{\rule{0.5em}{0ex}}f\left(x\right)=\frac{{x}^{2}-1}{{x}^{3}-4x}$

 Notes: (a) $y=\frac{{x}^{2}-1}{{x}^{3}-4x}=\frac{\left(x+1\right)\left(x-1\right)}{x\left({x}^{2}-4\right)}=\frac{\left(x+1\right)\left(x-1\right)}{x\left(x+2\right)\left(x-2\right)}$ (b) If $x=1$ then $y=0$. (c) If $x=-1$ then $y=0$. (d) If $x={2}^{+}$ then $y\to \infty$. $\left(\frac{\text{pos}.\text{pos}}{\text{pos}.\text{pos}.\text{pos}}\right)$ (e) If $x={2}^{-}$ then $y\to -\infty$. $\left(\frac{\text{pos}.\text{pos}}{\text{pos}.\text{pos}.\text{neg}}\right)$ (f) If $x={0}^{+}$ then $y\to \infty$. $\left(\frac{\text{pos}.\text{neg}}{\text{pos}.\text{pos}.\text{neg}}\right)$ (g) If $x={0}^{-}$ then $y\to -\infty$. $\left(\frac{\text{pos}.\text{neg}}{\text{neg}.\text{pos}.\text{neg}}\right)$ (h) If $x=-{2}^{+}$ then $y\to \infty$. $\left(\frac{\text{neg}.\text{neg}}{\text{neg}.\text{pos}.\text{neg}}\right)$ (i) If $x=-{2}^{-}$ then $y\to -\infty$. $\left(\frac{\text{neg}.\text{neg}}{\text{neg}.\text{neg}.\text{neg}}\right)$ (j) $y=\frac{{x}^{2}-1}{{x}^{3}-4x}\phantom{\rule{0.5em}{0ex}}\text{is the same as}\phantom{\rule{0.5em}{0ex}}\left(-y\right)=\frac{{\left(-x\right)}^{2}-1}{{\left(-x\right)}^{3}-4\left(-x\right)}$ So if we slip $y$ to $-y$ and $x$ to $-x$ the graph stays the same. (k) If $x\to \infty$ then $y\to {0}^{+}$. (l) If $x\to -\infty$ then $y\to {0}^{-}$.

Example:$\phantom{\rule{1em}{0ex}}$ $\text{Graph}\phantom{\rule{0.5em}{0ex}}\sqrt{x}+\sqrt{y}=1$.

 Notes: (a) If we switch $x$ and $y$ this graph stays the same. (b) If $x=0$ then $\sqrt{y}=1,$ so $y={1}^{2}=1$. (c) If we switch $y=0$ then $x=1$. (d) If $x=y$ then $\sqrt{x}+\sqrt{x}=1$ and $\sqrt{x}=\frac{1}{2},x=\frac{1}{4}$. (e) This graph should be similar to ${x}^{2}+{y}^{2}=1$ or $x+y=1$

Example:$\phantom{\rule{1em}{0ex}}$ $\text{Graph}\phantom{\rule{0.5em}{0ex}}\frac{{x}^{2}-1}{{x}^{2}+1}=f\left(x\right)$.

 Notes: (a) $y=\frac{{x}^{2}-1}{{x}^{2}+1}=\frac{{x}^{2}+1-2}{{x}^{2}+1}=1-\frac{2}{{x}^{2}+1}$

Example:$\phantom{\rule{1em}{0ex}}$ $\text{Graph}\phantom{\rule{0.5em}{0ex}}\mathrm{ln}\left(4-{x}^{2}\right)=f\left(x\right)$.

 Notes: (a) $y=\mathrm{ln}\left(4-{x}^{2}\right)=\mathrm{ln}\left(\left(2+x\right)\left(2-x\right)\right)=\mathrm{ln}\left(2+x\right)+\mathrm{ln}\left(2-x\right)$. (b) If we flip $x$ to $-x$ the graph stays the same since $y=\mathrm{ln}\left(4-{x}^{2}\right)=\mathrm{ln}\left(4-{\left(-x\right)}^{2}\right)$

$y=\mathrm{ln}\left(4-{x}^{2}\right)=\mathrm{ln}\left(2+x\right)+\mathrm{ln}\left(2-x\right)$.

Example:$\phantom{\rule{1em}{0ex}}$ $\text{Graph}\phantom{\rule{0.5em}{0ex}}y={x}^{\frac{2}{3}}{\left(6-x\right)}^{\frac{1}{3}}$

 Notes: (a) If $x=0$ then $y=0$. (b) If $x=6$ then $y=0$. (c) If $x\to \infty$ then $y\to {x}^{\frac{2}{3}}{\left(-x\right)}^{\frac{1}{3}}=-x$. (b) If $x\to -\infty$ then $y\to \infty \phantom{\rule{1em}{0ex}}\left(y\to -x\phantom{\rule{0.5em}{0ex}}\text{again}\right)$.

Example:$\phantom{\rule{1em}{0ex}}$ $\text{Graph}\phantom{\rule{0.5em}{0ex}}y=f\left(x\right)\phantom{\rule{0.5em}{0ex}}\text{when}\phantom{\rule{0.5em}{0ex}}x=\mathrm{cos}2\theta \phantom{\rule{0.5em}{0ex}}\text{and}\phantom{\rule{0.5em}{0ex}}y=\mathrm{cos}\theta$.

 Notes: $\mathrm{cos}2\theta ={\mathrm{cos}}^{2}\theta -{\mathrm{sin}}^{2}\theta ={\mathrm{cos}}^{2}\theta -\left(1-{\mathrm{cos}}^{2}\theta \right)=2{\mathrm{cos}}^{2}\theta -1$

$\text{So this problem is really asking for the graph of}\phantom{\rule{0.5em}{0ex}}\mathrm{cos}2\theta ={\mathrm{cos}}^{2}2\theta -1\phantom{\rule{0.5em}{0ex}}\text{when}\phantom{\rule{0.5em}{0ex}}x=\mathrm{cos}2\theta \phantom{\rule{0.5em}{0ex}}\text{and}\phantom{\rule{0.5em}{0ex}}y=\mathrm{cos}\theta$
$\phantom{\rule{0.5em}{0ex}}\text{i.e.}\phantom{\rule{1em}{0ex}}x=2{y}^{2}=1$

$\text{If}\phantom{\rule{0.5em}{0ex}}x=2{y}^{2}-1\text{.}\phantom{\rule{0.5em}{0ex}}\text{Then}\phantom{\rule{0.5em}{0ex}}x+1=2{y}^{2}\text{.}\phantom{\rule{0.5em}{0ex}}\text{So}\phantom{\rule{0.5em}{0ex}}{y}^{2}=\frac{x+1}{2}\text{.}\phantom{\rule{0.5em}{0ex}}\text{So}\phantom{\rule{0.5em}{0ex}}y=\sqrt{\frac{x+1}{2}}\text{.}\phantom{\rule{0.5em}{0ex}}\text{So}\phantom{\rule{0.5em}{0ex}}f\left(x\right)=\sqrt{\frac{x+1}{2}}$

## Notes and References

These are a typed copy of lecture notes given by Arun Ram on October 13, 2000.